Q. 10 - Calculus

Question

Evaluating iterated integrals: Sketch the region determined by the limits of the given iterated integrals, and then evaluate the integrals. 

${\int }_0^2{\int }_0^{y}x\sqrt{y^2-x^2} dx dy$

Step-by-Step Solution

Answer

${\int }_0^2{\int }_0^{y}x\sqrt{y^2-x^2} dx dy=\frac{128}{21}$

Step 1: Draw the region

The region determined by the limits of the given iterated integral is shown below,

Step 2: Evaluate the given integral

$$\begin{gathered} I={\int }_0^{y}x\sqrt{y^2-x^2} dx \\ \mathrm{Substitute}, \\ y^2-x^2=u^2 \\ \mathrm{Differentiate} \mathrm{w}.\mathrm{r}.\mathrm{t}. x \\ -2xdx=2udu \\ dx=-\frac{u}{x}du \end{gathered}$$

$$\begin{gathered} I=-{\int }_0^{y}x\sqrt{u^2} \frac{u}{x}du \\ I=-{\int }_0^y{u}^{2}du \\ I=-\left[\frac{u^3}{3}\right] \\ I=-{\left[\frac{{\left(y^2-x^2\right)}^3}{3}\right]}_0^y \\ I=\frac{y^6}{3} \end{gathered}$$

Therefore,

$$\begin{gathered} {\int }_0^2\left[{\int }_0^{y}x\sqrt{y^2-x^2} dx\right] dy=\frac{1}{3}{\int }_0^2{y}^6 dy \\ =\frac{1}{3}{\left[\frac{y^7}{7}\right]}_0^2 \\ =\frac{128}{21} \end{gathered}$$